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# How to solve for time given distance and acceleration using calculus?

Suppose you drop a ball from the top of a building that is 100 feet tall. When you drop the ball, you do not give it any initial velocity. Using calculus, how long does it take (in seconds) for the ball to hit the ground? Assume the acceleration due to gravity is 32.2 ft/sec^2

I'm not sure how to start. I know that acceleration is the derivative of velocity, velocity is the derivative of position, and so forth.

### 7 Answers

- Anonymous1 decade agoFavorite Answer
Since this is a Calculus problem, we will start with

dV/dt = a

where

dV/dt = acceleration of the ball as it is dropped from top of the building

a = acceleration due to gravity = 32.2 ft/sec^2 (constant)

The above equation then becomes

dV/dt = 32.2

V = ⌠32.2(dt)

and integrating the above,

V = 32.2t + C

where

C = integration constant

Since the ball was dropped at t = 0, then

0 = 0 + C and C = 0, hence the equation for V becomes

V = 32.2t

Since V = dS/dt, then

dS/dt = 32.2t

and

S = ⌠(32.2t)dt

Integrating the above,

S = (1/2)(32.2t^2) + C

and when t = 0, the ball has not moved yet, so

0 = 0 + C

and the above equation for "S" simplifies to

S = 16.1t^2

When S = 100 (distance travelled by the ball),

100 = 16.1t^2

and solving for "t"

t = 2.49 sec

Hope this helps.

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- vijLv 61 decade ago
D= Vi(t) + 1/2 a(t^2) In your question D = 100 ft, Vi = 0,

a = 32.2 ft/ sec^2

Sub the values in the equation and solve for t.

100 = 0(t) + 1/2(32.2) t^2

100 = 16.1 x t^2

100/ 16.1 = t^2

6.2 = t^2

2.5 = t

So ball takes 2.5 sec to hit the ground.

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- 1 decade ago
Well, the position equation for this is a*t^2 + v*t - h0, where h0 is the initial height, a is acceleration, and v is initial velocity. So, this position equation is 32.2*t^2 - 100. You set this equal to zero because you want the time that the height (or position) is 0 ft off the ground. So...

0 = 32.2*t^2 - 100

100 = 32.2*t^2 Divide by 32.2, take the (positive) square root, and you have the time when it hits the ground

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- SaraLv 44 years ago
Hi he.goes.the.distance Use the formula v^-u^2 = -2as (=2as when it is acceleration) v= 0 a = 16 s = 200 o^2 - u^2 = 2*16*200 u = 80 ft/sec Shy

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- hsueh010Lv 71 decade ago
a = dv/dt = -32.2 ft/s²

so integrate

v = -32.2ft/s² t + v(0)

but you said v(0) = 0 so

v = (-32.2ft/s²) t

but v = ds/dt

so

ds/dt = (-32.2ft/s²) t

s = (-32.2ft/s²)(1/2) t² + s(0)

so you want to know when s = 0 (final value on ground) when s(0) (initial position)= 100ft

0 = (-16.1ft/s²) t² + 100ft

Solve for t!

[-100/-16.1]s² = t²

6.2 s² = t²

t = 2.489 sec

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- grunfeldLv 71 decade ago
Hint: if you drop a ball regardless of height, what is your initial velocity?

s = -16.1t^2 + 100

- 100 = -16.1t^2

100 / 16.1 = t^2

10 / sqrt(16.1) = t

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- 1 decade ago
Use the equation:

distance = a*t^2 / 2 = g*t^2 / 2

t = √(2*distance/g)

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